兩個迴圈O(N^2)超時了
A = [4, 3, 2, 6] 長度n=4
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
探究F(0)和F(1) 關係
F(1)=(3* 6) + (0 * 4) + (1 * 3) + (2 * 2)-(3* 6)+(1* 4)+(1* 3)+(12)=F(0)+(1 * 6)+(1

class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        if not A:
            return 0
        n=len(A)
        res=0
        mysum=0
        for i in range(n):
            res+=i*A[i]
            mysum+=A[i]
        pre=res
        for j in range(n-1,0,-1):
            pre=pre+mysum-n*A[j]
            res=max(res,pre)
        return res